Question

If $\int \frac{\log \left(t+\sqrt{1+t^2}\right)}{\sqrt{1+t^2}}dt=\frac{1}{2}\left(g{\left(t\right)}^2\right)+C,$ where $C$ is a constant, then $g(2)$ is equal to :

• A. $2\log \left(2+\sqrt{5}\right)$
• B. $\log \left(2+\sqrt{5}\right)$
• C. $\frac{1}{\sqrt{5}}\log \left(2+\sqrt{5}\right)$
• D. $\frac{1}{2}\log \left(2+\sqrt{5}\right)$

Answer 1

Answer: (B)

$I=\int \frac{\log \left(t+\sqrt{1+t^2}\right)}{\sqrt{1+t^2}}dt$
$\because \frac{d}{dt}\left(\log \left(t+\sqrt{1+t^2}\right)\right)=\frac{1}{\sqrt{1+t^2}}$
$\Rightarrow I=\frac{1}{2}[\log {\left(t+\sqrt{1+t^2)}\right]}^2+C$
$\Rightarrow g(t)=\log \left(t+\sqrt{1+t^2}\right)$
$\Rightarrow g(2)=\log (2+\sqrt{5})$